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1 year ago

Can someone please help me with this

Mathematics

2 answers:

omeli [17]1 year ago

5 0

Answers:

x = 3
CD = 21
DE = 16
CE = 21

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Explanation:

The congruent base angles are D and E. Opposite those angles are the sides CE and CD. These opposite sides are the same length.

CE = CD

16x-27 = 4x+9

16x-4x = 9+27

12x = 36

x = 36/12

x = 3

This x value then leads to the following:

CD = 4x+9 = 4*3+9 = 12+9 = 21
DE = 7x-5 = 7*3-5 = 21-5 = 16
CE = 16x-27 = 16*3-27 = 48-27 = 21

We see that CD and CE are both 21 units long, which helps confirm we have the correct x value.

Lana71 [14]1 year ago

3 0

Answer:

x=3

CD= 21

DE= 16

CE= 21

Step-by-step explanation:

First, let's solve for x. Since angles D and E are congruent, you know that the length of lines CD and CE are equal. So, let's set up the equation by putting lines CD and CE equal to each other.

CD=CE

4x+9=16x-27

x= 3

Now that we've solved for x, we can now plug in the value into the other expressions to find the side lengths.

CD=4x+9

CD=4(3)+9

CD= 21

CE=16x-27

CE=16(3)-27

CE= 21

DE= 7x-5

DE= 7(3)-5

DE= 16

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gulaghasi [49]

The bearing of the plane is approximately 178.037°. $\blacksquare$

<h3>Procedure - Determination of the bearing of the plane</h3><h3 />

Let suppose that <em>bearing</em> angles are in the following <em>standard</em> position, whose vector formula is:

$\vec r = r\cdot (\sin \theta, \cos \theta)$ (1)

Where:

$r$ - Magnitude of the vector, in miles per hour.
$\theta$ - Direction of the vector, in degrees.

That is, the line of reference is the $+y$ semiaxis.

The <em>resulting</em> vector ( $\vec v$ ), in miles per hour, is the sum of airspeed of the airplane ( $\vec v_{A}$ ), in miles per hour, and the speed of the wind ( $\vec v_{W}$ ), in miles per hour, that is:

$\vec v = \vec v_{A} + \vec v_{W}$ (2)

If we know that $v_{A} = 239\,\frac{mi}{h}$ , $\theta_{A} = 180^{\circ}$ , $v_{W} = 10\,\frac{m}{s}$ and $\theta_{W} = 53^{\circ}$ , then the resulting vector is:

$\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})$

$\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right]$

Now we determine the bearing of the plane ( $\theta$ ), in degrees, by the following <em>trigonometric</em> expression:

$\theta = \tan^{-1}\left(\frac{v_{x}}{v_{y}} \right)$ (3)

$\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)$

$\theta \approx 178.037^{\circ}$

The bearing of the plane is approximately 178.037°. $\blacksquare$

To learn more on bearing, we kindly invite to check this verified question: brainly.com/question/10649078

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2 years ago

Solve -6f + 13 = 2f - 11

igor_vitrenko [27]

Answer:

f = 3

Step-by-step explanation:

Solve for f:

13 - 6 f = 2 f - 11

Subtract 2 f from both sides:

13 + (-6 f - 2 f) = (2 f - 2 f) - 11

-6 f - 2 f = -8 f:

-8 f + 13 = (2 f - 2 f) - 11

2 f - 2 f = 0:

13 - 8 f = -11

Subtract 13 from both sides:

(13 - 13) - 8 f = -13 - 11

13 - 13 = 0:

-8 f = -13 - 11

-13 - 11 = -24:

-8 f = -24

Divide both sides of -8 f = -24 by -8:

(-8 f)/(-8) = (-24)/(-8)

(-8)/(-8) = 1:

f = (-24)/(-8)

The gcd of -24 and -8 is -8, so (-24)/(-8) = (-8×3)/(-8×1) = (-8)/(-8)×3 = 3:

Answer: f = 3

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